A layer of ice having parallel sides floats on water. If lightis incident on the

Question

A layer of ice having parallel sides floats on water. If lightis incident on the upper surface of the ice at an angle ofincidence of 35.0°, what is the angle of refraction in thewater? I believe you have to start with the angle of incidence beingthe able in the air and the refraction angle of that would be inthe ice because it says the incident is on the upper surface. So this would be: 1 sin(35) = 1.309 sin (x) ..solving for therefraction angle in the ice x. I think from there you wouldhave to find the refracted angle into the water from the refractedangle in the ice: 1.309 sin(x) = 1. 333 sin(z) , z being therefracted angle in the water. If anyone has any hints or knows ifI'm going about this in the right way that would be great! Thanksfor any and all help! A layer of ice having parallel sides floats on water. If lightis incident on the upper surface of the ice at an angle ofincidence of 35.0°, what is the angle of refraction in thewater? I believe you have to start with the angle of incidence beingthe able in the air and the refraction angle of that would be inthe ice because it says the incident is on the upper surface. So this would be: 1 sin(35) = 1.309 sin (x) ..solving for therefraction angle in the ice x. I think from there you wouldhave to find the refracted angle into the water from the refractedangle in the ice: 1.309 sin(x) = 1. 333 sin(z) , z being therefracted angle in the water. If anyone has any hints or knows ifI'm going about this in the right way that would be great! Thanksfor any and all help!

Explanation / Answer

Given that The refractive index of ice (n1) =1.309 And the refractive index of water (n2) =1.333 The upper surface of the ice at an angle of incidence of(1) = 35.0° From the law of refraction we get that          n1sin1 =n2sin2 Now substitute all the above values to get the requiredanswer Now substitute all the above values to get the requiredanswer

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